Home
Class 12
PHYSICS
In a refrigerator, the low temperature c...

In a refrigerator, the low temperature coil of evaporator is at `-23^@C` and the compressed gas in the condenser has a temperature of `77^@C`. The amount of electrical energy spent in freezing 1 kg of water at `0^@C` is

A

`134400 J`

B

`1344 J`

C

`80000 J`

D

`3200 J`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of calculating the amount of electrical energy spent in freezing 1 kg of water at 0°C in a refrigerator with the given parameters, we can follow these steps: ### Step 1: Calculate the heat required to freeze 1 kg of water The heat required to freeze water can be calculated using the formula: \[ Q_L = m \cdot L_f \] where: - \( Q_L \) = heat required to freeze the water - \( m \) = mass of the water = 1 kg = 1000 grams - \( L_f \) = latent heat of fusion of water = 80 cal/g Substituting the values: \[ Q_L = 1000 \, \text{g} \cdot 80 \, \text{cal/g} = 80000 \, \text{cal} \] ### Step 2: Convert calories to joules To convert calories to joules, we use the conversion factor \( 1 \, \text{cal} = 4.2 \, \text{J} \): \[ Q_L = 80000 \, \text{cal} \cdot 4.2 \, \text{J/cal} = 336000 \, \text{J} \] ### Step 3: Determine the temperatures in Kelvin We need to convert the temperatures from Celsius to Kelvin: - Lower temperature (evaporator) = \(-23°C + 273 = 250 \, K\) - Higher temperature (condenser) = \(77°C + 273 = 350 \, K\) ### Step 4: Calculate the efficiency of the refrigerator The efficiency (COP) of the refrigerator can be calculated using the formula: \[ \text{COP} = \frac{T_L}{T_H - T_L} \] Substituting the values: \[ \text{COP} = \frac{250}{350 - 250} = \frac{250}{100} = 2.5 \] ### Step 5: Relate heat absorbed and work done Using the relationship: \[ Q_H = Q_L + W \] where \( W \) is the work done. We can express \( W \) in terms of \( Q_L \): \[ W = Q_H - Q_L \] ### Step 6: Calculate the heat rejected to the higher temperature body Using the efficiency: \[ W = \frac{Q_L}{\text{COP}} = \frac{336000 \, \text{J}}{2.5} = 134400 \, \text{J} \] ### Final Answer Thus, the amount of electrical energy spent in freezing 1 kg of water at 0°C is: \[ \boxed{134400 \, \text{J}} \]
Promotional Banner